Question

if x=2+√2 and y=2-√2,then the value of(x^2+y^2) is​

Answer 1

Answer:

${x}^{2} = (2 + \sqrt{2} )^{2} \\ = 4 + 2 + 4 \sqrt{2} = 6 + 4 \sqrt{2} \\ {y}^{2} = (2 - \sqrt{2} )^{2} \\ = 4 + 2 - 4 \sqrt{2} = 6 - 4 \sqrt{2} \\ {x}^{2} + {y}^{2} = 6 + 4 \sqrt{2} + 6 - 4 \sqrt{2} \\ = 12$

Answer 2

$x=2+\sqrt{2}$

$y=2-\sqrt{2}$

$x^{2}+ y^{2}$

$(2+\sqrt{2})^{2}$ $+ (2-\sqrt{2})^{2}$

$(4+2) + (4-2)$

$(a+b) (a-b) =a^{2}- b^{2}$

$4^{2} - 2^{2}$

$16 - 4$

$12$

∴Answer is $12$

hope it helps u